Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $t \neq 0$. $x = \dfrac{t + 5}{-2t^2 + 10t + 100} \div \dfrac{t^2 - 2t}{t^3 + 2t^2 - 8t} $
Solution: Dividing by an expression is the same as multiplying by its inverse. $x = \dfrac{t + 5}{-2t^2 + 10t + 100} \times \dfrac{t^3 + 2t^2 - 8t}{t^2 - 2t} $ First factor out any common factors. $x = \dfrac{t + 5}{-2(t^2 - 5t - 50)} \times \dfrac{t(t^2 + 2t - 8)}{t(t - 2)} $ Then factor the quadratic expressions. $x = \dfrac {t + 5} {-2(t + 5)(t - 10)} \times \dfrac {t(t - 2)(t + 4)} {t(t - 2)} $ Then multiply the two numerators and multiply the two denominators. $x = \dfrac {(t + 5) \times t(t - 2)(t + 4) } { -2(t + 5)(t - 10) \times t(t - 2)} $ $x = \dfrac {t(t - 2)(t + 4)(t + 5)} {-2t(t + 5)(t - 10)(t - 2)} $ Notice that $(t + 5)$ and $(t - 2)$ appear in both the numerator and denominator so we can cancel them. $x = \dfrac {t(t - 2)(t + 4)\cancel{(t + 5)}} {-2t\cancel{(t + 5)}(t - 10)(t - 2)} $ We are dividing by $t + 5$ , so $t + 5 \neq 0$ Therefore, $t \neq -5$ $x = \dfrac {t\cancel{(t - 2)}(t + 4)\cancel{(t + 5)}} {-2t\cancel{(t + 5)}(t - 10)\cancel{(t - 2)}} $ We are dividing by $t - 2$ , so $t - 2 \neq 0$ Therefore, $t \neq 2$ $x = \dfrac {t(t + 4)} {-2t(t - 10)} $ $ x = \dfrac{-(t + 4)}{2(t - 10)}; t \neq -5; t \neq 2 $